Suppose that we have the signal $$x(t) = e^{j\omega t} + e^{j\frac{3}{2} \omega t},$$ and we want to find a Fourier Series representation for that signal. Is this possible? According to my understanding, since the ratio of the frequencies of the two complex exponentials above is a rational number, the signal is periodic. Since it is periodic, it is possible to represent the signal as $$x(t) = \sum_{k = -\infty}^{\infty} a_k e^{jk\omega_0t},$$ where $\omega_0$ is the fundamental frequency of the signal?
1 Answer
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Yes, that's easily possible. The fundamental frequency of the given signal is $\omega_0=\omega/2$, and its Fourier series coefficients are
$$a_k=\begin{cases}1,\quad k\in\{2,3\}\\0,\quad\text{otherwise}\end{cases}$$
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1$\begingroup$ What about the substitution $\omega_0 = \omega / 4$? We would then have $x(t) = e^{j 4 \omega_0 t} + e^{j 6 \omega_0 t}$. Is this valid as well? $\endgroup$– user33568Feb 5, 2018 at 23:08
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$\begingroup$ Oh but in this case (the substitution in my comment above) $\omega_0$ does not equal the fundamental frequency of the signal. Am I correct? I find it a bit counter intuitive that although the signal is periodic with frequency $\omega$ it is only made up of higher frequency terms, since usually in a Fourier Series the coefficient $a_1$ is nonzero. $\endgroup$– user33568Feb 6, 2018 at 0:22